On the Brunn-Minkowski inequality for general measures with applications to new isoperimetric type-inequalities

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1 On the Brunn-Minkowski inequality for general measures with applications to new isoperimetric type-inequalities Galyna Livshyts, Arnaud Marsiglietti, Piotr Nayar, Artem Zvavitch April 24, 215 Abstract We show that for any product measure µ = µ 1 µ n on R n, where µ i have symmetric unimodal densities, the inequality µ(λa + ()B) 1/n λµ(a) 1/n + ()µ(b) 1/n holds true for any non-empty ideals A, B R n Moreover, using log-brunn-minkowski type inequalities due to C Saroglou, [S2], and to D Cordero-Erausquin, M Fradelizi and B Maurey, [CFM], we prove the above inequality for any ideals A, B and unconditional log-concave measures, as well as for general convex symmetric sets and even log-concave measures on the plane In addition, we deduce 1 n-concavity of the parallel volumes t µ(a+tb), Brunn's type theorem and certain analogues of Minkowski rst inequality We also provide examples showing optimality of our assumptions As a consequence of the above result, the Gaussian Brunn-Minkowski inequality holds true in the case of unconditional convex set, as well as in the case of general symmetric sets on the plane This partially solves the conjecture proposed by R Gardner and the fourth named author in [GZ] 21 Mathematics Subject Classication Primary 52A4; Secondary 6G15 Key words and phrases Convex bodies, Gaussian measure, Brunn-Minkowski inequality, ideals, Minkowski rst inequality, S-inequality, Brunn's theorem, Gaussian isoperimetry, log- Brunn-Minkowski inequality 1 Introduction The classical Brunn-Minkowski inequality states that for any two non-empty compact sets A, B in R n and any λ [, 1] we have vol n (λa + ()B) 1/n λ vol n (A) 1/n + () vol n (B) 1/n, (1) with equality if and only if B = aa + b, where a > and b R n Here vol n stands for the Lebesgue measure on R n and A + B = {a + b : a A, b B} 1

2 is the Minkowski sum of A and B Due to homogeneity of the volume, this inequality is equivalent to vol n (A + B) 1/n vol n (A) 1/n + vol n (B) 1/n The Brunn-Minkowski inequality turns out to be a powerful tool In particular, it easily implies the classical isoperimetric inequality: for any compact set A R n we have vol n (A t ) vol n (B t ), where B is an Euclidean ball satisfying vol n (A) = vol n (B) and A t stands for the t-enlargement of A, ie, A t = A + tb2 n, where B2 n is the unit Euclidean ball, B2 n = {x : x = 1} To see this it is enough to observe that vol n (A + tb n 2 ) 1/n vol n (A) 1/n + vol n (tb n 2 ) 1/n = vol n (B) 1/n + vol n (tb n 2 ) 1/n = vol n (B + tb n 2 ) 1/n Taking t + one gets a more familiar form of isoperimetry: among all sets with xed volume the surface area vol + vol n (A + tb2 n ) vol n (A) n ( A) = lim inf t + t is minimized in the case of the Euclidean ball We refer to [G] for more information on Brunn- Minkowski-type inequalities Using the inequality between means one gets an a priori weaker dimension free form of (1), namely vol n (λa + ()B) vol n (A) λ vol n (B) 1 λ (2) In fact (2) and (1) are equivalent To see this one has to take à = A/ vol n(a) 1/n, B = B/ vol n (B) 1/n and λ = λ vol n (A) 1/n /(λ vol n (A) 1/n +(1 λ) vol n (B) 1/n ) in (2) This phenomenon is a consequence of homogeneity of the Lebesgue measure The above notions can be generalized to the case of the so-called s-concave measures Here we assume that s (, 1], whereas in general the notion of s-concave measures makes sense for any s [, ] We say that a measure µ on R n is s-concave if for any non-empty compact sets A, B R n we have µ(λa + ()B) s λµ(a) s + ()µ(b) s (3) Similarly, a measure µ is called log-concave (or -concave) if for any compact A, B R n we have µ(λa + ()B) µ(a) λ µ(b) 1 λ (4) Let us assume that the support of our measure is non degenerate, ie, is not contained is any ane subspace of R n of dimension less than n It turns out that, in this case, there is a very useful description of log-concave and s-concave measures due to Borell, see [B]: a measure µ is log-concave if and only if it has a density of the form ϕ = e V, where V is convex (and may attain value + ) Such functions are called log-concave Moreover, µ is s-concave with s (, 1/n) if and only if it has a density ϕ such that ϕ s 1 sn is concave In the case s = 1/n the density has to satisfy the strongest condition ϕ(λx + ()y) max(ϕ(x), ϕ(y)) An example of such measure is the uniform measure on a convex body K R n Let us also notice that a measure with non-degenerate support cannot be s-concave with s > 1 It can be seen by taking n à = εa, B = εb, sending ε + and comparing the limit with the Lebesgue measure Inequality (2) says that Lebesgue measure is log-concave, whereas (1) means that it is also 1/n-concave In general log-concavity does not imply s-concavity for s > Indeed, consider the standard Gaussian measure γ n on R n, ie, a measure with density (2π) n/2 exp( x 2 /2) This density is clearly log-concave and therefore γ n satises (4) To see that γ n does not satisfy (3) for s > it suces to take B = {x} and send x Then the left hand side converges 2

3 to while the right hand side stays equal to λµ(a) s, which is strictly positive for λ > and µ(a) > One might therefore ask whether (3) holds true for γ n if we restrict ourselves to some special class of subsets of R n In [GZ] R Gardner and the fourth named author conjectured (Question 71) that γ n (λa + ()B) 1/n λγ n (A) 1/n + ()γ n (B) 1/n (5) holds true for any closed convex sets with A B and λ [, 1] and veried this conjecture in the following cases: (a) when A and B are products of intervals containing the origin, (b) when A = [ a 1, a 2 ] R n 1, where a 1, a 2 > and B is arbitrary, (c) when A = ak and B = bk where a, b > and K is a convex set, symmetric with respect to the origin It is interesting to note that the case (c) is related to the so-called Banaszczyk's theorem (Btheorem), [CFM] It states that for any convex symmetric set K the function t γ n (e t K) is log-concave Moreover, the same is true for any unconditional log-concave measures and unconditional sets It is an open question (called B-conjecture) whether this property is satised for any centrally symmetric log-concave measures and centrally symmetric convex sets This question has an armative answer for n = 2 due to the works of Livne Bar-on [Li] and of Saroglou [S2] In [S2] the proof is done by linking the problem to the new logarithmic-brunn- Minkowski inequality of Böröczky, Lutwak, Yang and Zhang, see [BLYZ1], [BLYZ2], [S1] and [S2] It turns out that the assertion of the B-conjecture for a measure µ and a symmetric convex body K formally implies the inequality µ(λak + ()bk) 1/n λµ(ak) 1/n + ()µ(bk) 1/n, ie, the Brunn-Minkowski inequality is satised for dilations of K, see [M2, Proposition 31] In [NT] T Tkocz and the third named author showed that in general (5) is false under the assumption A B For suciently small ε > and α < π/2 suciently close to π/2 the pair of sets A = {(x, y) R 2 : y x tan α}, B = {(x, y) R 2 : y x tan α ε} serves as a counterexample The authors however conjectured that (5) should be true for (centrally) symmetric convex bodies A, B Through this note K is some family of sets closed under dilations, ie, A K implies ta K for any t > In particular, we assume that for any K K we have K A general form of the Brunn-Minkowski inequality can be stated as follows Denition 1 We say that a Borel measure µ on R n satises the Brunn-Minkowski inequality in the class of sets K if for any A, B K and for any λ [, 1] we have µ(λa + ()B) 1/n λµ(a) 1/n + ()µ(b) 1/n (6) Before we state our results, we introduce some basic notation and denitions 3

4 Denition 2 1 We say that a function f : R n [, ) is unconditional if it satises the following two properties (i) For any choice of signs ε 1,, ε n { 1, 1} and any x = (x 1,, x n ) R n we have f(ε 1 x 1,, ε n x n ) = f(x) (ii) For any 1 i n and any x 1,, x i 1, x i+1,, x n R the function is non-increasing on [, ) t f(x 1,, x i 1, t, x i+1,, x n ) 2 A set A R n (which is not necessarily a product set) is called an ideal if 1 A is unconditional In other words, s set A R n is called unconditional if (x 1,, x n ) A implies (ε 1 x 1,, ε n x n ) A for any choice of signs ε 1,, ε n { 1, 1} The class of all ideal (in R n ) will be denoted by K I 3 A set A R n is called symmetric if A = A The class of all symmetric convex sets in R n will be denoted by K S 4 A measure µ on R n is called unconditional if it has an unconditional density Note that if an unconditional measure µ is a product measure, ie µ = µ 1 µ n, then the measures µ i are unconditional on R Our rst theorem reads as follows Theorem 1 Let µ be an unconditional product measure on R n Then µ satises the Brunn- Minkowski inequality in the class K I of all ideal in R n As we already mentioned, in such inequality for the measure with non-degenerate support, the power 1/n is the best possible In addition, it will be shown in Examples 1 and 2, in the end of the paper, that neither the assumption that µ is a product measure, nor the unconditionality of our sets A and B can be dropped Our strategy is to prove a certain functional version of (6) A functional version of the classical Brunn-Minkowski inequality is called the Prekopa-Leindler inequality, see [G] for the proof Theorem 2 (Prekopa-Leindler inequality, [P], [Le]) Let f, g, m be non-negative measurable functions on R n and let λ [, 1] If for all x, y R n we have m(λx + ()y) f(x) λ g(y) 1 λ then ( ) λ ( 1 λ m dx f dx g dx) Here we prove an analogue of the above inequality for unconditional product measures and unconditional functions f, g, m 4

5 Proposition 1 Fix λ, p (, 1) Suppose that m, f, g are unconditional Let µ be an unconditional product measure on R n Assume that for any x, y R n we have m(λx + ()y) f(x) p g(y) 1 p Then m dµ [ (λ p ) p ( ) ] 1 p n ( 1 p ) p ( f dµ g dµ) 1 p The above proposition allows us to prove the following lemma, which is in fact a reformulation of Theorem 1 Lemma 1 Let A, B be ideals in R n and let µ be an unconditional product measure on R n Then for any λ [, 1] and p (, 1) we have [ (λ ) p ( ) ] 1 p n µ(λa + ()B) µ(a) p µ(b) 1 p p 1 p It is worth noticing that the factor on the right hand side of this inequality replaces is some sense the lack of homogeneity of our measure µ The main idea of the proof is to introduce an additional parameter p λ and do the optimization with respect to p In the second part of this article we provide a link between the Brunn-Minkowski inequality and the logarithmic-brunn-minkowski inequality To state our observation we need two denitions Denition 3 Let K be a class of subsets closed under dilations We say that a family = ( λ ) λ [,1] of functions acting on K K (and having values in the class of measurable subsets of R n ) is a geometric mean if for any A, B K the set A λ B is measurable, satises an inclusion A λ B λa + ()B, and for any s, t > there is (sa) λ (tb) = s λ t 1 λ (A λ B) Denition 4 We say that a Borel measure µ on R n satises the logarithmic-brunn-minkowski inequality in the class of sets K with a geometric mean, if for any sets A, B K and for any λ [, 1] we have µ(a λ B) µ(a) λ µ(b) 1 λ, Remark 1 In what follows we use two dierent geometric means First one is a geometric mean S : K s K s K s, dened by the formula A S λ B = {x R n : x, u h λ A(u)h 1 λ B (u), u Sn 1 } Here h A is a support function of A, ie, h A (u) = sup x A x, u The second mean I : K I K I K I is dened by A I λ B = [ x 1 λ y 1 1 λ, x 1 λ y 1 1 λ ] [ x n λ y n 1 λ, x n λ y n 1 λ ] x A,y B It is straightforward to check, with the help of the inequality a λ b 1 λ λa+(1 λ)b, a, b, that both means are indeed geometric 5

6 In the Section 3 we prove the following fact Proposition 2 Suppose that a Borel measure µ with a radially decreasing density f, ie density satisfying f(tx) f(x) for any x R n and t [, 1], satises the logarithmic-brunn- Minkowski inequality, with a geometric mean, in a certain class of sets K Then µ satises the Brunn-Minkowski inequality in the class K In [S2] C Saroglou proved the logarithmic-brunn-minkowski inequality with geometric mean S for measures with even log-concave densities (that is, densities of the form e V, where V is even, ie V (x) = V ( x), and convex) in the plane R 2, in the class of symmetric convex sets (see Corollary 33 therein) Thus, as a consequence of Proposition 2 and Remark 1, we get the following corollary Corollary 1 Let µ be a measure on R 2 with an even log-concave density Then µ satises the Brunn-Minkowski inequality in the class K s of all symmetric convex sets in R 2 Moreover, in [CFM] (Proposition 8) the authors proved the following theorem (see also Proposition 42 in [S1]) Theorem 3 The logarithmic-brunn-minkowski inequality holds true with the geometric mean I for any measure with unconditional log-concave density in the class K I of ideal in R n We recall the argument in Section 3 As a consequence, applying our Proposition 2 together with Remark 1, we deduce the following fact Corollary 2 Let µ be an unconditional log-concave measure on R n Brunn-Minkowski inequality in the class K I of all ideals in R n Then µ satises the Let us now briey describe some corollaries of the Brunn-Minkowski type inequality we established, which are analogues to well-known osprings of Brunn-Minkowski inequality for the volume In what follows a pair (K, µ) is called nice if one of the following three cases holds (a) K = K I and µ is an unconditional product measure on R n (b) K = K I and µ is an unconditional log-concave measure on R n (c) K = K s and µ is an even log-concave measure on R 2 Corollary 3 Suppose that a pair (K, µ) is nice Let A, B K be convex Then the function t µ(a + tb) 1/n is concave on [, ) Indeed, for any λ [, 1] and t 1, t 2 we have µ(a + (λt 1 + ()t 2 )B) 1/n = µ((λ(a + t 1 B) + ()(A + t 2 B) 1/n λµ(a + t 1 B) 1/n + ()µ(a + t 2 B) 1/n Note that in the rst line we have used the convexity of A and B If B = B2 n is the Euclidean ball, the expression µ(a + tb) is called the parallel volume and has been studied in the case of Lebesgue measure by Costa and Cover in [C] as an analogue of concavity of entropy power in information theory The authors conjectured that for any measurable set A the parallel volume in 1 -concave In [FM] it has been shown that in general this conjecture is falls The authors n 6

7 proved the conjecture on the real line They also showed that it holds true on the plane in the case of connected sets In a recent paper [M] the second named author investigated the parallel volumes µ(a + tb n 2 ) in the context of s-concave measures He proved that t µ(a + t[ 1, 1]) 1/s is concave for s-concave measures on the real line However, as we already mentioned, even for the Lebesgue measure this result cannot be generalized to the higher dimensions Our Corollary 3 gives the Costa-Cover conjecture for any convex set A K, where (K, µ) is a nice pair Moreover, the Euclidean ball B n 2 can be replaced with any convex set B K Second, we may state the following analogue of the Brunn's theorem on volumes of sections of convex bodies Corollary 4 Suppose that a pair (K, µ) is nice Let K K be a convex set Then the function 1 t µ(a {x 1 = t}) is -concave on its support n 1 To prove this let us take K t = K {x 1 = t} By convexity of K we get λk t1 + ()K t2 K λt1 +(1 λ)t 2 Thus, using (6), for any λ [, 1] and t 1, t 2 R such that K t1 and K t2 are both non-empty, we get µ(k λt1 +(1 λ)t 2 ) 1 n 1 µ(λkt1 + ()K t2 ) 1 n 1 λµ(kt1 ) 1 n 1 + ()µ(kt2 ) 1 n 1 Third, let us mention the relation of our result to the Gaussian isoperimetric inequality, Ehrhard's inequality and the so-called S-inequality Let us begin with the Gaussian isoperimetry It turns out that for any measurable set A R n and any t >, the quantity γ n (A t ) is minimized, among all sets with prescribed measure, for the half spaces H a,θ = {x R n : x, θ a}, with a R and θ S n 1 This is a famous result established by Sudakov and Tsirelson, [ST], as well as independently by Borell, [B2], and can be seen as one of the cornerstones of the concentration of measure theory Innitesimally, it says that among all sets with prescribed measure the half spaces are those with the smallest Gaussian measure of the boundary, ie, the quantity γ + n (A) = lim inf t + γ n (A + tb2 n ) γ n (A) t A more general statement is the so-called Ehrhard's inequality It says that for any non-empty measurable sets A, B we have Φ 1 (γ n (λa + ()B)) λφ 1 (γ n (A)) + ()Φ 1 (γ n (B)), λ [, 1], (7) where Φ(t) = γ 1 ((, a]) This inequality has been considered for the rst time by Ehrhard in [E], where the author proved it assuming that both A and B are convex Then Lataªa in [L] generalized Erhard's result to the case of arbitrary A and convex B In it's full generality, the inequality (7) has been established by Borell, [B3] Note that our inequality (5), valid for unconditional ideals, is an inequality of the same type, with Φ(t) replaced with t n None of them is a direct consequence of the other However, unlike the Gaussian Brunn-Minkowski inequality, the Ehrhard's inequality (in fact a more general form of it where λ and are replaced with α and β, under the conditions α + β 1 and α β 1) gives the Gaussian isoperimetry as a simple consequence Let us also describe the S-inequality of Lataªa and Oleszkiewicz, see [LO] It states that for any t > 1 and any symmetric convex body K the quantity γ n (tk) is minimized, among all subsets with prescribed measure, for the strips of the form S L = {x R n : x 1 L} This 7

8 result admits an equivalent innitesimal version, namely, among all symmetric convex bodies K with prescribed Gaussian measure the strips S L minimize the quantity d γ dt n(tk) t=1, which is equivalent to maximizing M(K) = x 2 dγ n (x), K see [KS] or [NT3] For a general measure µ with a density e ψ, one can show that the innitesimal version of S-inequality is an issue of maximizing the quantity M µ (K) = x, ψ dµ(x), (8) K see equation (11) below Not much is known about this general problem In the unconditional case it has been solved for some particular product measures like products of Gamma and Weibull distributions, see [NT2] It turns our that inequality (5) implies a certain mixture of Gaussian isoperimetry and reverse S-inequality Namely, we have the following Corollary Corollary 5 Let A, B be two ideals in R n (or general symmetric convex sets in R 2 ) and let r > Then we have rγ + n (A) + M(A) nγ n (rb n 2 ) 1 n γn (A) 1 1 n with equality for A = rb n 2 Let us state and prove a more general version of Corollary 5 Let µ + (A) be the µ boundary measure of A, ie, µ + µ(a + tb2 n ) µ(a) (A) = lim inf t + t Moreover, let us introduce the so-called rst mixed volume of arbitrary sets A and B, with respect to measure µ Namely Clearly, µ + (A) = nv µ 1 (A, B n 2 ) V µ 1 (A, B) = 1 lim inf n t + µ(a + tb) µ(a) t Corollary 6 Let A, B K and suppose that (K, µ) is a nice pair Then we have In particular, V µ 1 (A, B) + 1 n M µ(a, B) µ(b) 1/n µ(a) 1 1/n (9) rµ + ( A) + M µ (A) nµ(rb n 2 ) 1/n µ(a) 1 1/n (1) To prove this we note that for any sets A, B K and any ε [, 1) we have ( ) 1/n A µ(a + εb) 1/n (1 ε)µ + εµ(b) 1/n 1 ε Indeed, it suces to use Theorem 1 with λ = 1 ε and à = A/(1 ε), B = B Note that for ε = we have equality Thus, dierentiating at ε = we get 1 n µ(a) 1 n 1 nv µ 1 (A, B) µ(b) 1 n µ(a) 1 n + 1 n µ(a) 1 n 1 d dt µ(ta) t=1 8

9 By changing variables we obtain d dt µ(ta) t=1 = d e ψ(tx) t n dx = nµ(a) dt t=1 Thus, A A x, ψ(x) dµ(x) = nµ(a) M µ (A) (11) µ(a) 1 n 1 V µ 1 (A, B) µ(b) 1 1 n n µ(a) 1 n 1 M µ (A) This is exactly (9) To get (1) one has to take B = rb2 n in (9) The above inequalities can be seen as an analogues of the so-called Minkowski rst inequality for the Lebesgue measure, see [G] and [Sn], which says that for any two convex bodies A, B in R n we have V voln 1 (A, B) vol n (A) 1 1 n voln (B) 1 n The rest of this article is organized as follows In the next section we rst show how Lemma 1 implies Theorem 1 Then Lemma 1 is deduced from Proposition 1 Finally, we prove Proposition 1 In Section 3 we prove Proposition 2 and recall the proof of Theorem 3 In the last section we give examples showing optimality of some of our results and state open questions 2 Proof of Theorem 1 We rst show how Lemma 1 implies Theorem 1 Proof of Theorem 1 Without loss of generality we can assume that λ (, 1) Let us assume for a moment that µ(a)µ(b) > Then we can use Lemma 1 with Note that Then p = λ p = λµ(a)1/n + ()µ(b) 1/n µ(a) 1/n, [ (λ p λµ(a) 1/n (, 1) (12) λµ(a) 1/n + ()µ(b) 1/n 1 p = λµ(a)1/n + ()µ(b) 1/n µ(b) 1/n ) p ( ) ] 1 p n µ(a) p µ(b) 1 p = ( λµ(a) 1/n + ()µ(b) 1/n) n 1 p Thus the inequality in Lemma 1 becomes µ(λa + ()B) ( λµ(a) 1/n + ()µ(b) 1/n) n Now suppose that, say, µ(b) = Since B is a non-empty ideal, we have B Therefore, λa λa + ()B Let ϕ µ be the unconditional density of µ Hence, µ(λa + ()B) µ(λa) = ϕ(x) dx = λ n ϕ(λy) dy λa A = λ n ϕ(λy 1,, λy n ) dy = λ n ϕ(λ y 1,, λ y n ) dy A A λ n ϕ( y 1,, y n ) dy = λ n µ(a) A 9

10 Therefore, µ(λa + ()B) 1/n λµ(a) 1/n = λµ(a) 1/n + ()µ(b) 1/n Next we show that Proposition 1 implies Lemma 1 Proof of Lemma 1 We can assume that λ (, 1) Let us take m(x) = 1 λa+(1 λ)b (x), f(x) = 1 A (x), g(x) = 1 B (x) Clearly, f, g and m are unconditional It is easy to verify that for any p (, 1) we have m(λx + ()y) f(x) p g(y) 1 p Our assertion follows from Proposition 1 For the proof of Proposition 1 we need a one dimensional Brunn-Minkowski inequality for unconditional measures Lemma 2 Let A, B be two symmetric intervals and let µ be unconditional on R Then for any λ [, 1] we have µ(λa + ()B) λµ(a) + ()µ(b) Proof We can assume that A = [ a, a] and B = [ b, b] for some a, b > Let ϕ µ be the density of µ Then our assertion is equivalent to λa+(1 λ)b ϕ µ (x) dx λ a ϕ µ (x) dx + () b ϕ µ (x) dx In other words, the function t t ϕ µ(x) dx should be concave on [, ) This is equivalent to t ϕ µ (t) being non-increasing on [, ) Proof of Proposition 1 We proceed by induction on n Let us begin with the case n = 1 We can assume that f, g > Indeed, if we multiply the functions m, f, g by numbers c m, c f, c g satisfying c m = c p f c1 p g, the hypothesis and the assertion do not change Therefore, taking c f = f 1, c g = g 1, c m = f p g (1 p) we can assume that f = g = 1 Then the sets {f > t} and {g > t} are non-empty for t (, 1) Moreover, λ{f > t} + (){g > t} {m > t} Indeed, if x {f > t} and y {g > t} then m(λx + ()y) f(x) p g(y) 1 p > t p t 1 p = t Thus, λx + ()y {m > t} Therefore, using Lemma 2, we get m dµ = λ 1 1 = λ = λ µ({m > t}) dt 1 µ({m > t}) dt µ(λ{f > t} + (){g > t}) dt µ({f > t}) dt + () µ({f > t}) dt + () f dµ + () g dµ 1 µ({g > t}) dt µ({g > t}) dt 1

11 Now, using the inequality pa + (1 p)b a p b 1 p we get λ f dµ + () g dµ = p λ f dµ + (1 p) g dµ (13) p 1 p ( ) p ( ) 1 p ( ) p ( 1 p λ f dµ g dµ) (14) p 1 p Next, we do the induction step Let us assume that the assertion is true in dimension n 1 Let m, f, g : R n [, ) be unconditional For x, y, z R we dene functions m z, f x, g y by m z (x) = m(z, x), f x (x) = f(x, x), g y (x) = g(y, x) Clearly, these functions are also unconditional Moreover, due to our assumptions on m, f, g we have m λx +(1 λ)y (λx + ()y) = m(λx + ()y, λx + ()y) f(x, x) p g(y, y) 1 p = f x (x) p g y (y) 1 p Let us decompose µ in the form µ = µ 1 µ, where µ 1 is a measure on R Note that µ 1 and µ are unconditional and µ is a product measure Thus, by our induction assumption we have [ (λ ) p ( ) ] 1 p n 1 ( ) p ( 1 p m λx +(1 λ)y d µ f x d µ g y d µ) (15) p 1 p Now we dene the functions [ (λ ) p ( ) ] 1 p (n 1) M(z ) = m z (ξ) d µ(ξ), (16) p 1 p F (x ) = f x (ξ) d µ(ξ), G(y ) = g y (ξ) d µ(ξ) (17) Using nequality (15) we immediately get that M(λx + (y )) F (x ) p G(y ) 1 p Moreover, it is easy to see that M, F, G are unconditional on R Thus, using Lemma 2 (the one-dimensional case), we get ( ) p ( ) 1 p ( ) p ( 1 p λ M(z) dµ 1 (z) F (x) dµ 1 (x) G(y) dµ 1 (y)) (18) p 1 p Observe that M(z) dµ 1 (z) = Similarly, Our assertion follows = [ (λ p [ (λ p F (x ) dµ 1 (x ) = ) p ( ) ] 1 p (n 1) 1 p ) p ( ) ] 1 p (n 1) 1 p f dµ, m z (ξ) dµ n 1 (ξ) dµ 1 (z ) m dµ G(y ) dµ 1 (y ) = g dµ 11

12 3 Proof of Proposition 2 In this section we rst prove Proposition 2 The argument has a avour of our previous proof Proof of Proposition 2 Let us rst assume that µ(a)µ(b) > From the denition of geometric mean we have A p B pa + (1 p)b, for any p (, 1) Thus, ( µ(λa + ()B) = µ p λ ) (( ) ( )) λ A + (1 p) p 1 p B µ p A p 1 p B ( (λ ) p ( ) 1 p = µ A p B) p 1 p ( ) p ( ) 1 p λ 1 λ Let t = p 1 p and C = A p B From the concavity of the logarithm it follows that t 1 We have µ(tc) = f(x) dx = t n f(tx) dx t n f(x) dx = t n µ(c) (19) Therefore, tc µ(λa + ()B) t n µ(a p B) t n µ(a) p µ(b) 1 p = Taking gives p = C C [ (λ λµ(a) 1/n λµ(a) 1/n + ()µ(b) 1/n µ(λa + ()B) 1/n λµ(a) 1/n + ()µ(b) 1/n If, say, µ(b) = then by (19) and the fact that B we get p ) p ( ) ] 1 p n µ(a) p µ(b) 1 p 1 p µ(λa + ()B) 1/n µ(λa) 1/n λµ(a) 1/n = λµ(a) 1/n + ()µ(b) 1/n We now sketch the proof of Theorem 3 Proof Let A, B K I and let us take f, g, m : R n + R + given by f = 1 A R n +, g = 1 B R n + and m = 1 (A I λ B) R Let n ϕ µ be an unconditional log-concave density of µ We dene + F (x) = f(e x 1, e xn )ϕ µ (e x 1, e xn )e x 1++x n, G(x) = g(e x 1, e xn )ϕ µ (e x 1, e xn )e x 1++x n, M(x) = m(e x 1, e xn )ϕ µ (e x 1, e xn )e x 1++x n One can easily check, using the denition of K I and the denition of the geometric mean I λ, as well as the inequality ϕ µ (e λx 1+(1 λ)y 1,, e λxn+(1 λ)yn ) = ϕ µ ((e x 1 ) λ (e y 1 ) 1 λ,, (e xn ) λ (e yn ) 1 λ ) ϕ µ (λe x 1 + ()e y 1,, λe xn + ()e yn ) ϕ µ (e x 1,, e xn ) λ ϕ µ (e y 1,, e yn ) 1 λ, that the functions F, G, M satisfy assumptions of Theorem 2 The rst inequality above follows from the fact that ϕ µ is unconditional As a consequence, we get µ((a I λ B) Rn +) µ(a R n +) λ µ(b R n +) 1 λ The assertion follows from unconditionality of our measure µ and the fact that A, B and A I λ B are ideals 12

13 4 Examples and open problems Example 1 The assumption, that our measure µ in Theorem 1 is product, is important Indeed, let us take the square C = { x, y 1} R 2 and take the measure with density ϕ(x) = C(x) C(x) This density is unconditional, however non product Let us dene ψ(a) = µ(ac) The assertion of Theorem 1 implies that ψ is concave However, we have ψ(a) = 2a for a [1, 2], which is strictly convex Thus, µ does not satisfy (6) Example 2 In general, under the assumption that our measure µ is unconditional and product, one cannot prove that Theorem 1 holds true for arbitrary symmetric convex sets To see this, let us take the product measure µ = µ µ on R 2, where µ has an unconditional density ϕ(x) = p + (1 p)1 [ 1/ 2,1/ 2] (x) for some p [, 1] This measure is not nite, however one can always restrict it to a cube [ C, C] 2 to get a nite product measure If C is suciently large the example given below also produces and example for the restricted measure To simplify the computation let us rotate the whole picture by angle π/4 Then consider the rectangle R = [ 1, 1] [ λ, λ] for λ 1/2 As in the previous example, it is enough to show that the function ψ(a) = µ(ar) is not concave Let us consider this function only on the interval [1/λ, ) The condition λ 1/2 ensures that the point (a, λa) lies in the region with density p 2 Let us introduce lengths l 1, l 2, l 3 (see the picture below) Note that l 1 = 2λa, l 2 = 2(λa 1) and l 3 = a (1 + λa) Let ω(a) = µ(ar) We have ω(a) = p l1 + l 2 + p 2 l1 2 + p 2 l p 2 l 3 λa 2 = 2 + 4p(2λa 1) + 2p 2 λ 2 a 2 + 2p 2 (λa 1) 2 + 4p 2 λa(a a) = 2(1 p) 2 + 4pλa(pa + 2 2p) = d + d 1 a + d 2 a 2, where d = 2(1 p) 2, d 1 = 8p(1 p)λ, d 2 = 4p 2 λ We show that ψ is strictly convex for p (, 1) and < λ < 1/2 Then ψ > is equivalent to 2ωω > (ω ) 2 But 2ω(a)ω (a) (ω (a)) 2 = 4d 2 (d + d 1 a + d 2 a 2 ) (2d 2 a + d 1 ) 2 = 4d 2 d d 2 1 = 32λp 2 (1 p) 2 64λ 2 p 2 (1 p) 2 = 32λp 2 (1 p) 2 (1 2λ) > (a, λa) l 3 l 1 l 2 (1, ) Let us state some open questions that arose during our study Question Let us assume that the measure µ has an even log-concave density (not-necessarily product) Does the assertion of Theorem 1 holds true for arbitrary symmetric sets A and B? If not, is it true under additional assumption that the sets are unconditional or that the measure is product? In particular, can one remove the assumption of unconditionality in the Gaussian Brunn-Minkowski inequality? 13

14 References [B] C Borell, Convex set functions in d-space, Period Math Hungarica 6 (1975), [B2] C Borell, The Brunn-Minkowski inequality in Gauss space, Invent Math 3, 2 (1975), [B3] C Borell, The Ehrhard inequality, C R Math Acad Sci Paris 337 (23), no 1, [BLYZ1] K J Böröczky, E Lutwak, D Yang and G Zhang, The log-brunn-minkowski inequality, Adv Math 231 (212), [BLYZ2] K J Böröczky, E Lutwak, D Yang and G Zhang, The logarithmic Minkowski problem, J Amer Math Soc 26 (213), ) [CFM] D Cordero-Erausquin, M Fradelizi, B Maurey, The (B) conjecture for the Gaussian measure of dilates of symmetric convex sets and related problems, J Funct Anal 214 (24), no 2, [C] M Costa, T M Cover, On the similarity of the entropy power inequality and the Brunn- Minkowski inequality, IEEE Trans Inform Theory 3 (1984), no 6, [E] A Ehrhard, Symétrisation dans l'espace de Gauss, Math Scand 53 (1983) [FM] M Fradelizi, A Marsiglietti, On the analogue of the concavity of entropy power in the Brunn-Minkowski theory, Adv in Appl Math 57 (214), 12 [G] RJ Gardner, The Brunn-Minkowski inequality, Bull Amer Math Soc (NS) 39 (22), no 3, [GZ] RJ Gardner, A Zvavitch, Gaussian Brunn-Minkowski inequalities, Trans Amer Math Soc 362 (21), no 1, [KS] S Kwapie«, J Sawa, On some conjecture concerning Gaussian measures of dilatations of convex symmetric sets, Studia Math 15 (1993), no 2, [L] R Lataªa, A note on the Ehrhard inequality, Studia Math 118 (1996) [Li] A Livne Bar-on, The (B) conjecture for uniform measures in the plane, Geometric Aspects of Functional Analysis, Lecture Notes in Mathematics Volume 2116, 214, [LO] R Lataªa, K Oleszkiewicz, Gaussian measures of dilatations of convex symmetric sets, Ann Probab 27 (1999), [Le] L Leindler, On a certain converse of Hölder's inequality, II, Acta Sci Math (Szeged) 33 (1972), [M] A Marsiglietti, Concavity properties of extensions of the parallel volume, arxiv: v3 [M2] A Marsiglietti, On the improvement of concavity of convex measures, arxiv:

15 [NT] P Nayar, T Tkocz, A note on a Brunn-Minkowski inequality for the Gaussian measure, Proc Amer Math Soc 141 (213), no 11, [NT2] P Nayar, T Tkocz, S-inequality for certain product measures, Math Nachr 287 (214), no 4, [NT3] P Nayar, T Tkocz, The unconditional case of the complex S-inequality, Israel J Math 197 (213), no 1, 9916 [P] A Prékopa, On logarithmic concave measures and functions, Acta Sci Math 34 (1973): pp [S1] C Saroglou, Remarks on the conjectured log-brunn-minkowski inequality, 214, Geom Dedicata (to appear) [S2] C Saroglou, More on logarithmic sums of convex bodies, arxiv: [Sn] R Schneider, Convex bodies: the Brunn-Minkowski theory, Second expanded edition Encyclopedia of Mathematics and its Applications, 151 Cambridge University Press, Cambridge, 214 xxii+736 pp ISBN: [ST] V N Sudakov, B S Cirel'son, Extremal properties of half-spaces for spherically invariant measures, Problems in the theory of probability distributions, II Zap Nau n Sem Leningrad Otdel Mat Inst Steklov (LOMI) 41 (1974), 1424, 165 Galyna Livshyts Department of Mathematical Sciences Kent State University Kent, OH 44242, USA address: glivshyt@kentedu Arnaud Marsiglietti Institute for Mathematics and its Applications University of Minnesota 27 Church Street, 432 Lind Hall, Minneapolis, MN USA address: arnaudmarsiglietti@imaumnedu Piotr Nayar Institute for Mathematics and its Applications University of Minnesota 27 Church Street, 432 Lind Hall, Minneapolis, MN USA address: nayar@imaumnedu 15

16 Artem Zvavitch Department of Mathematical Sciences Kent State University Kent, OH 44242, USA address: 16